Thursday, November 15 ∗ Solutions ∗ Parametrizations and Integrals1. Consider the ellipsoid with implicit equationx2a2+y2b2+z2c2=1.(a) Parametrize this ellipsoid.Solution. One could use the parametrizationx = a sinφcos θ, y =b sinφ sinθ, z =c cos φ, 0 ≤φ ≤π, 0 ≤θ ≤2π.(b) Set up, but do not evaluate, a double integral that computes its surface area.Solution. Since r(φ,θ) =〈a sinφcosθ, b sinφsinθ, c cosφ〉, one hasrφ=〈a cosφcosθ, b cos φsin θ, −c sinφ〉, rθ=〈−a sinφsinθ, b sinφcosθ, 0〉,sorφ×rθ=〈bc sin2φcos θ, ac sin2φsin θ, ab sinφcosθ〉.Therefore|rφ×rθ|=qb2c2sin4φcos2θ +a2c2sin4φsin2θ +a2b2sin2φcos2φ,and the surface area is computed byArea =Z2π0Zπ0|rφ×rθ|dφdθ=Z2π0Zπ0qb2c2sin4φcos2θ +a2c2sin4φsin2θ +a2b2sin2φcos2φ dφdθ.2. Letr(u, v) =〈(2 +cos u) cos v,(2 +cosu)sin v,sin u〉,where 0 ≤u ≤2π and 0 ≤v ≤2π.(a) Sketch the surface parametrized by this function.Solution. The sketch of the surface is as follows.(b) Compute its surface area.Solution. By the parametrization, one hasru=〈−sin u cos v, −sinu sin v, cosu〉,rv=〈−(2 +cos u)sin v, (2 +cosu)cosv, 0〉,and soru×rv=〈−(2 +cos u)cos u cos v, −(2 +cos u)cos u sin v, −(2 +cos u)sin u〉 .Therefore |ru×rv|=2 +cos u, and the surface area is computed byArea =Z2π0Z2π0|ru×rv|dud v =Z2π0Z2π0(2 +cos u) dud v =8π2.3. Consider the surface integralÏΣz dSwhere Σ is the surface with sides S1given by the cylinder x2+y2=1, S2given by the unit disk inthe x y-plane, and S3given by the plane z =x +1. Evaluate this integral as follows:(a) Parametrize S1using (θ, z) coordinates.Solution. One can parametrize S1byx =cosθ, y =sin θ, z = z, 0 ≤θ ≤2π, 0 ≤z ≤cosθ +1.(b) Evaluate the integral over the surface S2without parametrizing.Solution. Since z =0 on S2, we knowÎS2z dS =0.(c) Parametrize S3in Cartesian coordinates and evaluate the resulting integral using polar co-ordinates.Solution. One can parametrize S3in Cartesian coordinatesx = x, y = y, z = x +1, −1 ≤ x ≤1, −p1 −x2≤ y ≤p1 −x2.Now we move to evaluate the integralÎΣz dS. ObviouslyÏΣz dS =ÏS1z dS +ÏS2z dS +ÏS3z dS := I1+I2+I3.To estimate I1, using the parametrization in (a), one hasr(θ, z) =〈cosθ , sin θ, z〉.Thenrθ=〈−sin θ, cosθ, 0〉, rz=〈0, 0, 1〉,andrθ×rz=〈cosθ, sinθ, 0〉.So |rθ×rz|=1, andI1=Z2π0Zcosθ+10z d zdθ =Z2π0(cosθ +1)22dθ=Z2π0cos2θ +2 cosθ +12dθ =3π2.In (b) we know I2=0. To evaluate I3, by the parametrization in (c), one hasr(x, y) =〈x, y, x +1〉, −1 ≤ x ≤1, −p1 −x2≤ y ≤p1 −x2,and sorx=〈1,0, 1〉, ry=〈0,1, 0〉, rx×ry=〈−1,0, 1〉.Thus |rx×ry|=p2, and the surface integral isI3=Z1−1Zp1−x2−p1−x2(x +1)p2 d ydx =Ïx2+y2≤1(x +1)p2 d ydx.To evaluate this integral, one can use the polar coordinatesx =r cosθ, y =r sinθ, 0 ≤r ≤1, 0 ≤θ ≤2π.Therefore,I3=Z2π0Z10(r cosθ +1)p2 rdr dθ =p2π.Adding up all three integrals, one getsÏΣz dS = I1+I2+I3=3π2+p2π.4. Let C be the circle in the plane with equation x2+y2−2x =0.(a) Parametrize C as follows. For each choice of a slope t, consider the line Ltwhose equationis y = t x. Then the intersection Lt∩C of Ltand C contains two points, one of which is(0,0). Find the other point of intersection, and call its x−and y−coordinates x(t) and y(t).Compute a formula for r(t ) =〈 x(t), y(t)〉. Check your answer with your TA.Solution. Bring y = t x into x2+ y2−2x = 0, then one has x2+t2x2−2x = 0, and it iseasy to get x =21+t2, and then y =2t1+t2. Thus r(t ) =〈21+t2,2t1+t2〉.(b) Suppose that t =pqis a rational number. Show that x(p/q) and y(p/q) are also rationalnumbers. Explain how, by clearing denominators in x(p/q)−1 and y(p/q), you can find aa triple of integers U ,V, and W for which U2+V2=W2.Solution. Plug t =pqinto the the parametrization, one getsx(p/q) =2q2p2+q2, y(p/q) =2pqp2+q2,and both of them are rational numbers. Since (x −1)2+y2=1, and x(p/q)−1 =q2−p2p2+q2, thenone hasµq2−p2p2+q2¶2+µ2pqp2+q2¶2=1.By settingU =q2−p2, V =2pq, W = p2+q2,one has U2+V2=W2.(c) ComputeZC12〈−y, x〉·dr using your parametrization above.Solution. Since r =〈21+t2,2t1+t2〉, one has r0=〈−4t(1+t2)2,2−2t2(1+t2)2〉. ThenZC12〈−y, x〉·dr =Z∞−∞12〈−2t1 +t2,21 +t2〉·〈−4t(1 +t2)2,2 −2t2(1 +t2)2〉 d t=Z∞−∞2(1 +t2)2d t
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